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If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,
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Updated on: 02 Nov 2012, 02:37
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If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x? A. 999 B. 1001 C. 1003 D. 1004 E. 1005
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Originally posted by ruturaj on 09 Jul 2011, 02:21.
Last edited by Bunuel on 02 Nov 2012, 02:37, edited 1 time in total.
Renamed the topic, edited the question and moved to PS forum.




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Re: Faster way to solve such questions
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09 Jul 2011, 03:42
ruturaj wrote: If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x? 999 1001 1003 1004 1005 I think the fastest way to do this is a simple understanding of mean . Sum of the deviations of the numbers in the set from the mean is always zero 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 mean is 999 so the list is 65321+2+2+3+5... this shud total to zero but this is 5 , hence we need a number that is 5 more than the mean to get a +5 and make it zero hence the answer is 999+ 5 1004 D




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Re: Faster way to solve such questions
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09 Jul 2011, 03:20
993 = 1000  7 994 = 1000  6 996 = 1000  4 997 = 1000  3 998 = 1000  2 1001 = 1000 + 1 1001 = 1000 + 1 1002 = 1000 + 2 1004 = 1000 + 4 So Sum of terms = (1000 * 9  22 + 8) + x Average = (9000  14 + x)/10 = 999 999 = 1000  1 So (9000  14 + x) = 10000  10 => x = 1000 + 4 = 1004 Answer  D
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Re: Faster way to solve such questions
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09 Jul 2011, 03:36
As avg is given we can divide them in 2 groups 1st is of elements below 999 and 2nd is ofelements abv 999 if we find now the diff that each element has with 999 that is (999each element) for the 1st group (eg. 999998=1,999997=2,....) and (each element999) for the 2nd group(eg. 1001999=2,1002999,....) now add the diff for each group separately
for 1st group sum is 17 and for 2nd group its 12
1712=5
so we need an element which is in the 2nd group and 5 more than 999 thats 1004
This method relies on the fact that avg is the middle point of a set from Weightage point of view.... I think this way it cab be solved within a min...



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Re: Faster way to solve such questions
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09 Jul 2011, 05:22
\(\frac{3 + 4 + 6 + 7 + 8 + 11 + 11 + 12 + 14 + x}{10} = 9\)
\(\frac{76 + x}{10} = 9\)
\(x = 90  76 = 14\)
\(990 + 14 = 1004\)



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Re: Faster way to solve such questions
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13 Jul 2011, 05:00
ruturaj wrote: If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x? 999 1001 1003 1004 1005 When I say the mean of the ages of a group is 25, it means that effectively every person's age can be replaced by 25 and the sum of ages will remain the same. The sum of deviation of all numbers from the mean is 0. Deviation means how much the number is more or less than the mean. If mean is 999, deviation of 993 from mean is 6. Deviation of 994 from mean is 5. The sum of all deviations except that of x is:  6  5  3  2  1 + 2+ 2 + 3 + 5 = 5 But this sum should be 0. Hence deviation of x from 999 should be +5. Therefore, x should be 1004.
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Re: Faster way to solve such questions
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01 Nov 2012, 22:59
ruturaj wrote: If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x? 999 1001 1003 1004 1005 This problem can be solved in 8 secs When you have sequences like this with very little differences we may deviate with signs and differences. To get rid of them and wasting time in adding big numbers, we can replace these big numbers with very small numbers.
Remember, every sec in GMAT is Invaluable
example: If 993(the first term) is assumed as 1, 994 > 993+1 is replaced by 1+1 ie 2, and so on so forth,then the sequence will be
Mean is 999 which is 993+6 ie 1+6=7 (1+2+4+5+6+9+9+10+12+x)/10=7 x=12, which is 1 + 11 = 993+11= 1004
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Re: If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,
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01 Oct 2014, 03:33
Mean = 999 Dividing the numbers in 2 groups; < 999 & > 999 Group I : < 999999 6 = 993 999 5 = 994 999 3 = 996 999 2 = 997 999 1 = 998 Total ve = 17Group II: > 999999 +2 = 1001 999 +2 = 1001 999 +3 = 1002 999 +5 = 1004 Total +ve = 12For the mean (including x) to be 999, total +ve should be equal to total ve So, 999+5 = 1004 Answer = D
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If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,
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14 Feb 2015, 05:59
Great tips! I did it somehow different:
993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 , x  M = 999
1000*4...=.4000 900*5.....=.4500 90*5.......=..450 +28........=....28 +8..........=.....8 Addition..=8986
\(\frac{8986+x}{10}\)= 999
x = 1004



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Re: If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,
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12 Feb 2018, 17:36
ruturaj wrote: If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x?
A. 999 B. 1001 C. 1003 D. 1004 E. 1005 Since the numbers are “pretty” large and each is “pretty” close to the average, without using the conventional formula (i.e., average = sum/quantity), we can determine the value of x by using the difference between each known number and the average. If we subtract 999, the average, from each number, we have: 6, 5, 3, 2, 1, 2, 2, 3, 5 We see that the sum of the differences is (17) + 12 = 5. To counter the 5 (i.e., 5 less than the average), we need a value that is 5 more than the average. That is, we need x to be 999 + 5 = 1004 so that the average of all values (including x) will be 999. Answer: D
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Re: If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, &nbs
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